The sum of the two digit numbers above is a three digit number PQ5, wh

ok, so we have
XY+
YX
= PQ5, where all the digits are distinct.
from this, we can conclude that X+Y is equal to either a 1 digit number or a 2 digit number, the units digit of which would be 5.

from the given possibilities:
I x = 7, then we must have Y equal to 8 so that the units digit would be 5.
78+87 = 165 - since all the digits are distinctive, then yes, X can be 7.

II x = 8, this works the same way as I. if x =8 then y = 7, and we get 87+78 = 165. again, this satisfies the conditions.
If we stumble and don't know what to do next, we can just eliminate A, B, and C, and try to make a smart guess with a chance of getting the right answer 1/2.

III x = 9, let's see, if X =9, then Y must be 6.
96+69 = 165. Same thing as I and II right? no! it is not. The condition is that X, Y, P, and Q are distinct digits, which mean that no digit should be repeated. In this case we have Q has the same digit as Y. This does not satisfy the condition, and thus, the answer is D.

D: I and II only.
Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater.
XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5
=> 1Q5 divisible by 11
=> Q = 1+5 = 6, PQ5 = 165
=> X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. But notice that X,Y,P,Q is different from each other so we cannot take 6 twice.
Conclusion : X e (7,8). Choice D.

QUESTION #1:

_X Y
+Y X
________

The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?

I) 7
II) 8
III) 9

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I , II and III




_X Y
+Y X
________


X+Y = 5

Possible values of x and Y are (X,Y)

(1,4)
(2,3)
(3,2)
(4,1)

The above pair can not produce 3 digit distinct numbers when they are added, hence we neglect these options.

(6,9)
(7,8)
(8,7)
(9,6)

The sum of the two digit numbers above is a three digit number PQ5

_X Y
+Y X
________

Check for the possible options -

(6,9)

69+96 = 165

(7,8)

78+87 = 165

(8,7)

87+78 = 165

(9,6)

96 + 69 = 165

Given the conditions only the folowing set of numbers product 3 digit distinct numbers in the form of PQ5 and the sum of its units digit is 5

(6,9)
(7,8)
(8,7)
(9,6)

So, the possible values of X can be 6,7,8 and 9.

Among the given options all 7 ,8 & 9 ca be the possible values of X.

Hence IMHO answer is (E) all 1,2 and 3.

JAMBOBREE OFFICIAL SOLUTION:

Since XY and YX are 2 digit numbers, there sum has to be less than 200. Thus, P = 1. Also sum of the digits X and Y should end with 5. Hence X + Y = 5 or 15 or 25… so that 5 is the units digit and tens digit is carried over. However, X + Y is less than 18 (as they are digits) and greater than 10 (else sum won’t be a 3-digit number). Thus, X + Y = 15, which gives
X Y
+ Y X
-------
P Q 5
=> 1 6 5

Alternatively: when a two digit number is added to another two digit number formed by reversing the digits of the first two digit number,the result is a multiple of 11. So in this case as PQ5 has to be a multiple of 11. So, using the divisibility rule of 11, P =1 and Q = 6.

Now, X+Y = 15,so all possible values of x and Y are
Case 1 X = 6 Y = 9
Case 2 X = 7 Y = 8
Case 3 X = 8 Y =7
Case 4 X =9 Y = 6

But as it is given that each letter X, Y, P, and Q represents a different non zero digit AND we know that Q = 6 so case 1 and case 4 are not possible

So the only possible values of x are 7 and 8

Hence the correct answer is option D.

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